Question

Let $f(x)=(x+\log_3 x)^2 + x^2 \forall x>0$, then derivative of $f^{-1}(x)$ w.r.t. $x$ at $x=2$ is

• A. $\frac{\ln 3}{4 \ln 3+2}$
• B. $\frac{1}{6 \ln 3+2}$
• C. $\frac{\ln 3}{3 \ln 3+8}$
• D. $\frac{3 \ln 3}{2 \ln 3-8}$

Answer 1

Answer: (A)

$\frac{\ln 3}{4 \ln 3+2}$

Answer 2

Let $y=f(x)$. We want to find $(f^{-1})'(2)$. The formula is $(f^{-1})'(y) = \frac{1}{f'(x)}$ where $y=f(x)$. First, find $x$ such that $f(x)=2$. $f(1) = (1+\log_3 1)^2 + 1^2 = (1+0)^2 + 1 = 1+1 = 2$. So, $f(1)=2$, which means $f^{-1}(2)=1$. Next, find the derivative of $f(x)$: $f'(x) = \frac{d}{dx} \left( (x+\log_3 x)^2 + x^2 \right)$. Using $\log_3 x = \frac{\ln x}{\ln 3}$, so $\frac{d}{dx}(\log_3 x) = \frac{1}{x \ln 3}$. $f'(x) = 2(x+\log_3 x) \cdot \left(1 + \frac{1}{x \ln 3}\right) + 2x$. Evaluate $f'(1)$: $f'(1) = 2(1+\log_3 1) \cdot \left(1 + \frac{1}{1 \cdot \ln 3}\right) + 2(1)$ $f'(1) = 2(1+0) \cdot \left(1 + \frac{1}{\ln 3}\right) + 2$ $f'(1) = 2 \left(\frac{\ln 3 + 1}{\ln 3}\right) + 2 = \frac{2 \ln 3 + 2 + 2 \ln 3}{\ln 3} = \frac{4 \ln 3 + 2}{\ln 3}$. Finally, $(f^{-1})'(2) = \frac{1}{f'(1)} = \frac{1}{\frac{4 \ln 3 + 2}{\ln 3}} = \frac{\ln 3}{4 \ln 3 + 2}$.